Uhmmm... Let's see if I can get it correct
x2(1 - x2004) + y2(1 - y2004) = 0
The sign of each term depends on (1 - x2004), and (1 - y2004) respectively.
Both can be 0, that's the case mentioned above.
The other case is one must be negative and one must be positive. Say (1 - x2004) < 0, and (1 - y2004) > 0
That means x > 1, or x < -1 (cannot satisfy either equation), and -1 < y < 1. Hence, this case cannot happen.

From the first original equations, we know that |x|,|y|<=1, and therefore, we have:
[tex]0\leq{x}^{2}(1-x^{2004}),0\leq{y}^{2}(1-y^{2004})[/tex]
since the sum of these terms equals 0, we must have:
[tex]x^{2}(1-x^{2004})=0[/tex]
And:
[tex]y^{2}(1-y^{2004})=0[/tex]
This system of equations is readily solved, and by insertion of the solutions of this sytem into the first system, we find which solutions that system has.